A particle starts its `SHM` from mean position at `t = 0`. If its time period is `T` and amplitude `A` then the distance travelled by the particle in the time from `t = 0 to t = (5T)/(A)` is

Options

A. `A`
B. `3A`
C. `4A`
D. `5A`

Explanation

To find the distance traveled by a particle executing Simple Harmonic Motion (SHM) from time \( t = 0 \) to \( t = \frac{5T}{4} \), we can follow these steps:

### Step 1: Understand the Motion
The particle starts from the mean position at \( t = 0 \). In SHM, the particle moves back and forth between the maximum amplitude \( A \) and the mean position. The time period of the motion is \( T \).

### Step 2: Determine the Motion in One Complete Cycle
In one complete cycle (time \( T \)), the particle travels from the mean position to the maximum amplitude \( A \), back to the mean position, to the negative amplitude \( -A \), and back to the mean position. The total distance traveled in one complete cycle is:
\[ \text{Distance in one cycle} = A + A + A + A = 4A \]

### Step 3: Calculate the Distance for \( t = 0 \) to \( t = T \)
From \( t = 0 \) to \( t = T \), the particle travels the entire distance of one complete cycle, which is \( 4A \).

### Step 4: Calculate the Distance for \( t = T \) to \( t = \frac{5T}{4} \)
Now, we need to find the distance traveled from \( t = T \) to \( t = \frac{5T}{4} \). The time interval from \( T \) to \( \frac{5T}{4} \) is:
\[ \Delta t = \frac{5T}{4} - T = \frac{5T}{4} - \frac{4T}{4} = \frac{T}{4} \]

In this time interval \( \Delta t = \frac{T}{4} \), the particle will move from the mean position to the maximum amplitude \( A \) and then back to the mean position.

### Step 5: Calculate the Distance in the Interval \( \Delta t \)
During this interval of \( \frac{T}{4} \), the particle travels:
- From the mean position (0) to the amplitude \( A \) (distance \( A \))
- Back to the mean position (distance \( A \))

Thus, the total distance traveled during this interval is:
\[ \text{Distance} = A + A = 2A \]

### Step 6: Total Distance Traveled
Now, we can sum up the distances traveled:
- Distance from \( t = 0 \) to \( t = T \): \( 4A \)
- Distance from \( t = T \) to \( t = \frac{5T}{4} \): \( 2A \)

Therefore, the total distance traveled by the particle from \( t = 0 \) to \( t = \frac{5T}{4} \) is:
\[ \text{Total Distance} = 4A + 2A = 6A \]

### Final Answer
The distance traveled by the particle in the time from \( t = 0 \) to \( t = \frac{5T}{4} \) is \( 6A \).

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